1 PHYS208 p-n junction January 15, 2010
List of topics (1) Density of states Fermi-Dirac distribution Law of mass action Doped semiconductors Dopinglevel p-n-junctions
1 Intrinsic semiconductors
List of topics Density of states The number of conduction electrons is equal to the number of holes. The density of states for the free electron in the conduction band is g c (E) = V 2π m e is the electron effective mass. E c is the energy at the conduction band edge. ( ) 2m 3/2 e h 2 (E E c ) 1/2. (2)
For the free holes in the valence band, the density of states is m h g v (E) = V 2π is the hole effective mass. E c is the energy at the valence band edge. ( ) 2m 3/2 h h 2 (E v E) 1/2. (3)
List of topics Fermi-Dirac distribution The probability of a particular energy state being occupied is given by f(e) = 1 exp ( ) (4) E µ kt + 1 µ is the chemical potential. T is the temperature. k is the Boltzmann constant.
In conduction band there are very few electrons. This means that f e (E) = 1 e E µ kt + 1 << 1 (5) It follows that that the exponential is large, so that which is true if 1 e E µ kt + 1 E µ e kt (6) E µ >> kt (7)
The concentration of electrons in the conduction band is n e = 1 V E c g c (E)f(E)dE = 2 ( m e kt 2π h 2 ) 3/2 e Ec µ kt (8) g c (E) = V 2π Go to the chemical potential ( ) 2m 3/2 e h 2 (E E c ) 1/2. (9)
For holes in valence band f h (E) = 1 f e (E) e E µ kt (10) f h is the probability that a valence orbital is occupied. g v (E) = V 2π The concentration of holes in the valence band is ( ) 2m 3/2 h h 2 (E v E) 1/2. (11) n h = 1 V Ev ( ) m 3/2 g v (E)f h (E)dE = 2 h kt 2π h 2 e Ev µ kt (12)
List of topics Law of mass action Since the number of holes is the same as the number of electrons in an intrinsic semiconductor n e = n h = n i (13) ( ) m 3/2 n e = 2 e kt 2π h 2 e Ec µ kt (14) ( ) m 3/2 n h = 2 h kt 2π h 2 e Ev µ kt (15) ( ) 3 kt n e (T )n h (T ) = 4 2π h 2 (m em h) 3/2 e Eg kt = n 2 i, (16)
List of topics n i is the intrinsic carrier density. E g = E c E v E g is the energy gap. ( ) 3/2 kt n i (T ) = 2 2π h 2 (m em h) 3/4 e Eg 2kT (17) The result is independent of µ and it drops exponentially at the band gap increases. The expression (16) is the law of mass action. Since n e (T ) = n i (T ), we obtain by equating (17) and (8) µ = E c 1 2 E g + 3 ( ) m 4 kt ln h m e (18)
List of topics Doped semiconductors (19) Impurity atoms with more valence electrons than the rest of the atoms in the crystall are called Donors and it forms n-type semiconductors. Impurity atoms with less valence electrons than the rest of the atoms in the crystall are called Acceptors and it forms p-type semiconductors. If the material to be doped is silicon (4 valence electrons), one can add phosphorus (5 valence electrons) to get a n-type semiconductor and one can add boron (3 valence electrons) to get a p-type semiconductor.
+4 Si +4 Si +4 Si +4 Si +4 Si +4 Si CONDUCTION ELECTRON HOLE -q +q +4 +5 Si P +4 Si +4 Si +3 B +4 Si +4 Si +4 Si +4 Si +4 Si +4 Si +4 Si (a) (b)
List of topics Doping level We want to show that the energy level for the extra electrons will be right below the conduction band for a n-doped semiconductor. As a model we use the Hydrogen Hamiltonian, where the electronic mass is replaced by the corresponding effective mass. H = h2 2 e2 2m e 4πɛ r ɛ 0 r (20) ɛ r is the relative dielectric constant since we are not in vacuum. We know that the energies in hydrogen are given by E n = m ( ) e e 2 2 1 2 h 2 4πɛ 0 n = 13.6 ev (21) 2 n2
m e = Cm e (22) Then the energy of the groundstate (n = 1) in our model is ( ) E model = m e e 2 2 2 h 2 = m ( ) e e 2 2 C C 4πɛ 0 ɛ r 2 h 2 = E 4πɛ 0 ɛ 2 hydrogen r ɛ 2 r Example (23) For phosphorus doping of silicon ɛ r = 11.7 and m e 0.2m e The binding energy in our model is then The energy of the doping states are E 0.02 ev (24) E d = E c + E (25) Since E < 0 the states are right below the edge of the conduction band. If we look at the holes in the same way as the electrons it can be shown that the energy level for the extra holes will be right above the edge of the valence band.
n-doped p-doped
f( ε, µ,t) ε µ k T + 1 e 1 E c µ E v
f( ε, µ,t) ε µ k T + 1 e 1 µ E c E v µ E v E c
f( ε, µ,t) ε µ k T + 1 e 1 µ E c E v µ E v E c
f( ε, µ,t) f( ε, µ,t) ε µ k T + 1 e 1 µ E c E v µ E v E c
f( ε, µ,t) f( ε, µ,t) 1 f( ε, µ,t) f( ε, µ,t) ε µ k T + 1 e 1 µ E c E v µ E v E c
List of topics 2 p-n junctions (26) In the p-doped and the n-doped materials the fermi levels will be at different energies.
From thermodynamics we know that the fermi levels must be the same if they are joined together.
When p-doped and n-doped materials are joined together, there is an inbalance between electrons and holes on each side. Electrons and holes diffuse diffustion current. Positive charge on n-side and negative on p-side potential drift current. At equilibrium 0 = j drift + j diff
The diffusion current is given by Fick s law, J = D n (27) where D is the diffusion constant and n is the density of carriers. Regarding the electrons we find that the charge current caused by diffusion is The current caused by the electric field j diff = ej = ed n (28) j drift = σe = enµ mob E, (29) where again n is the carrier density and µ mob is the mobility. At equilibrium 0 = j drift + j diff = en(r)µ mob E + ed n(r) (30) Assuming that there only is change in the x-direction, the equation becomes n(x)µ mob E x (x) = D dn(x) dx, (31) which gives us the electrical field in terms of the electron distribution: E(x) = D 1 dn µ mob n dx. (32)
Interface between the n-doped and the p-doped semiconductor is very depleted of charge carriers. The resistance there will be much larger than in the rest of the semiconductor. Practically all of the potential fall will be in this region. Thus V c = Φ( ) Φ( ) = = D n( ) 1 dn µ mob n( ) n(x) dx dx = E x dx D µ mob n( ) n( ) dn n = D µ mob ln n( ) n( ).
Far from the interface the carrier densities are that of p-doped and n-doped semiconductors, respectively. Since most of the electrons introduced by the impurities are free at normal temperatures, n( ) = N D. Use Law of mass action (16) to find n( ) N A = n 2 i (T ) on p-side. Thus n( ) n( ) = N DN A n 2 i (T ). To establish the relationship between D and µ consider a constant strong E x giving the equation D dn dx = µ mobe x n(x), which has solution n(x) e µ mobe xx/d. (33) But from Boltzmann statistics we have n(x) e eexx/k BT. (34)
n(x) e eexx/k BT. (35) n(x) e µ mobe xx/d. (36) Since these two exponentials must be the same to ensure consistency, we find that D = k BT µ mob e, (37) which is the Einstein-Nernst equation. Upon substitution Example V c = k BT e Si crystal: N A = N D = 10 16 cm 3. ln N DN A. (38) n 2 i (T ) kt e 0.025 V. For Si when T = 300 K, n i 10 10 cm 3. Then V c = 0.72 V (39)
Assume ρ(x) has a constant value in a region of thickness d p and another constant value in a region of thickness d n. Everywhere else we assume ρ(x) = 0. Assume all electrons from the region d n fill all the holes in the region d p. N A d p = N D d n.
From Poisson s equation we know that de x /dx = 4πρ(x). Thus for d p < x < 0 E x (x) = 4πeN A (d p + x), (40) and for 0 < x < d n E x (x) = 4πeN D (d n x). (41) Outside the depletion zone we assume there is no electric field.
Since E x = dφ/dx this can be realised by 0 x < d p 2πeN Φ(x) = A (d p + x) 2 d p < x < 0 V c 2πeN D (d n x) 2 (42) 0 < x < d n V c x > d n
Continuity at x = 0 gives 2πeN A d 2 p = V c 2πeN D d 2 n. Defining d = d p + d n, and using d n N D = d p N A we can write d n = N A N A + N D d, d p = N D N A + N D d. Inserting this for d n and d p we find ( NA ND 2 V c = 2πe + N ) DNA 2 d 2 = 2πed 2 N A N D, (N A + N D ) 2 N A + N D or solving for d, d = Vc 2πe NA + N D N A N D. (43)
It is very difficult to influence the drift current since the electric field is very large at the junction. It is possible to change the diffusion current by changing the density of electrons and holes. At V external = 0 the current over the p-n junction is j = j diff j drift = j 0 j 0 = 0. If V external 0 j drift will not be much changed, but in the diffution current there will be an extra Boltzmann factor e ev/k BT. Thus the current with an external field is j = (e ev/k BT 1)j 0. (44) At V > 0 there is a current growing exponentially as we increase V, and at large negative values of V we are left with only the drift current which is very small.